What is the extreme f(x)=sqrt(5-x^2) ?

The extreme f(x)=sqrt(5-x^2) is Minimum(-sqrt(5),0),Maximum(0,sqrt(5)),Minimum(sqrt(5),0)

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extreme f(x)=sqrt(5-x^2)

extreme points

f (x) = 5 - x^{2}

Minimum (- 5, 0), Maximum (0, 5), Minimum (5, 0)

Solution steps

f^{'} (x) = - \frac{x}{5 - x ^{2}}

Find intervals:

Increasing

: - 5 < x < 0,

Decreasing

: 0 < x < 5

Plug

x = - 5

into

5 - x^{2} : 0

Minimum (- 5, 0)

Plug

x = 0

into

5 - x^{2} : 5

Maximum (0, 5)

Plug

x = 5

into

5 - x^{2} : 0

Minimum (5, 0)

Minimum (- 5, 0), Maximum (0, 5), Minimum (5, 0)

f (x) = x - x

f (x) = (\frac{ln ( x )}{x ^{2}})

f (x) = x^{3} - 4 x^{2} - 11 x + 30

f (x) = x^{2} - 8 x + 16

f (x) = \frac{1}{16 - x ^{2} - y ^{2}}

What is the extreme f(x)=sqrt(5-x^2) ?
The extreme f(x)=sqrt(5-x^2) is Minimum(-sqrt(5),0),Maximum(0,sqrt(5)),Minimum(sqrt(5),0)